Question

Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 245000 kg and a velocity of 0.315 m/s in the horizontal direction, and the second having a mass of 57500 kg and a velocity of -0.125 m/s in the horizontal directionWhat is their final velocity, in meters per second?

Answer 1

Answer:

+0.231 m/s

Explanation:

The problem can be solved by using the law of conservation of momentum. In fact, we have that the total momentum before the collision must be equal to the total momentum after the collision:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$

where we have

m1 = 245000 kg is the mass of the first car

m2 = 57500 kg is the mass of the second car

u1 = 0.513 m/s is the initial velocity of the first car

u2 = -0.125 m/s is the initial velocity of the second car

v = ? is the final velocity of the two cars together, after the collision

Solving the equation for v, we find

$v=\frac{m_1 u_1 + m_2 u_2}{m_1 +m_2}=\frac{(245000 kg)(0.315 m/s)+(57500 kg)(-0.125 m/s)}{245000 kg+57500 kg}=+0.231 m/s$

And the direction (positive sign) is the same as the initial direction of the first car.

Answer 2

Answer:

The net velocity in horizontal direction: $v=0.2314\ m.s^{-1}$

Explanation:

Given:

• mass of cart 1, $m_1=245000 \ kg$

• mass of cart 2, $m_2=57500 \ kg$

• velocity of cart 1, $v_1=0.315 \ m.s^{-1}$

• velocity of cart 2, $v_2=-0.125 \ m.s^{-1}$

Using the law of conservation of momentum when the two carts are coupled together.

We know momentum is given as:

$p=m.v$

Net momentum:

$m.v = m_1.v_1+m_2.v_2$

$m.v=245000 \times 0.315 +57500 \times (-0.125)$

$m.v=69987.5\ kg.m.s^{-1}$ ...................................(1)

∵Net mass:

$m=m_1+m_2$

$m=245000 +57500$

$m=302500\ kg$

Now from  equation (1) the net velocity in horizontal direction:

$302500\times v=69987.5$

$v=0.2314\ m.s^{-1}$